Q46 / 160 · AP EAMCET 2020 (18 Sep Shift 2) · Mathematics
Let \(M=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) and \(N=\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]\). Then \(N M^{10} N^{-1}=\)
- Correct answerA\(\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]\)
- B\(\left[\begin{array}{cc}1 & -5 \\ 0 & 1\end{array}\right]\)
- C\(\left[\begin{array}{cc}1 & -10 \\ 0 & 1\end{array}\right]\)
- D\(\left[\begin{array}{cc}1 & 10 \\ 0 & 1\end{array}\right]\)
Solution
\(\begin{aligned}
M & =\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] \\
M^2 & =\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] \\
M^2 & =\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right] \\
M^3 & =\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right] \\
& \vdots \\
M^{10} & =\left[\begin{array}{ll}
1 & 10 \\
0 & 1
\end{array}\right] \\
N & =\left[\begin{array}{ll}
1 & 0 \\
0 & 2
\end{array}\right] \\
|N| & =2-0=2 \\
\operatorname{Adj}^N & =\left[\begin{array}{ll}
2 & 0 \\
0 & 1
\end{array}\right] \\
N^{-1} & =\frac{1}{|N|} \text { Adj A } \\
N^{-1} & =\frac{1}{2}\left[\begin{array}{ll}
2 & 0 \\
0 & 1
\end{array}\right] \\
N^{-1} & =\left[\begin{array}{ll}
1 & 0 \\
0 & \frac{1}{2}
\end{array}\right]
\end{aligned}\)
Consider,
\(\begin{aligned}
\mathbf{N} \cdot \mathbf{M}^{10} \cdot \mathbf{N}^{-1} & =\left[\begin{array}{ll}
1 & 0 \\
0 & 2
\end{array}\right]\left[\begin{array}{cc}
1 & 10 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 0 \\
0 & \frac{1}{2}
\end{array}\right] \\
& =\left[\begin{array}{cc}
1 & 10 \\
0 & 2
\end{array}\right]\left[\begin{array}{cc}
1 & 0 \\
0 & \frac{1}{2}
\end{array}\right]=\left[\begin{array}{ll}
1 & 5 \\
0 & 1
\end{array}\right]
\end{aligned}\)
Hence, option (a) is correct.
Hence, the correct answer is option: 'A'.